Question

$\rm \implies a {x}^{2} - 5x + 2 = 0$
Solve for a such that the equation has exactly 1 root.

Give proper explanation. ​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the Concept of Quadratic Equations has been used. Here we see that we are given a quadratic equation. We know that a Quadratic Equation has 2 roots either can be equal, imaginary or distinct but here we have to go with equal case since, we need to find the value of a such that there are only one roots that is only one value of x.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{For\;equal\;roots\;:\;b^{2}\;-\;4ac\;=\;\bf{0}}}$

_______________________________________________

★ Solution :-

Given,

✒ ax² - 5x + 2 = 0

Here, a = a, b = -5 and c = 2.

We know that,

$\\\;\;\sf{:\mapsto\;\;For\;equal\;roots\;:\;b^{2}\;-\;4ac\;=\;\bf{0}}$

$\\\;\;\sf{:\mapsto\;\;b^{2}\;-\;4ac\;=\;\bf{0}}$

By applying values, we get,

$\\\;\;\sf{:\mapsto\;\;(-\:5)^{2}\;-\;4\;\times\;a\;\times\;2\;=\;\bf{0}}$

$\\\;\;\sf{:\mapsto\;\;25\;-\;8\;\times\;a\;=\;\bf{0}}$

$\\\;\;\sf{:\mapsto\;\;\;-\;8\;\times\;a\;=\;\bf{-\:25}}$

$\\\;\;\sf{:\mapsto\;\;\;8\;\times\;a\;=\;\bf{25}}$

$\\\;\;\underline{\underline{\bf{:\mapsto\;\;\;a\;=\;\bf{\dfrac{25}{8}}}}}$

Also, when a = 0 , we get,

✒ ax² - 5x + 2 = 0

✒ 0x² - 5x + 2 = 0

✒ -5x = -2

✒ 5x = 2

✒ x = 2/5

This also, gives single root of equation since it forms linear equation. Thus,

a = 0

is also correct answer.

$\\\;\underline{\boxed{\tt{Hence\;\;for\;\;single\;\;roots,\;a\;=\;\bf{\dfrac{25}{8}\;\;\tt{or}\;\;\bf{0}}}}}$

_______________________________________________

★ More to know :-

• Verification ::

We need to verify, if the equation has single roots or not for the value of a we got. Then,

✒ ax² - 5x + 2 = 0

✒ (25/8)x² - 5x + 2 = 0

✒ 25x² - 40x + 8 = 0

Now using the Quadratic Formula, we get,

$\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:b\;\pm\;\sqrt{b^{2}\;-\;4ac}}{2a}}$

$\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:40\;\pm\;\sqrt{(-\:40)^{2}\;-\;4\;\times\;25\;\times\;8}}{2\;\times\;25}}$

$\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:(-\:40)\;\pm\;\sqrt{1600\;-\;1600}}{50}}$

$\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{40\;\pm\;\sqrt{0}}{50}}$

$\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{40}{50}\;=\;\dfrac{4}{5}}$

Clearly,

$\\\;\bf{\Rightarrow\;\;x\;=\;\dfrac{4}{5}\;or\;\dfrac{4}{5}}$

This is because, this is Quadratic equation so it must have two equal roots.

This gives that this equation has only single root that is 4/5 when a = 25/8

So our answer is correct.

Hence, Verified.

_______________________________________________

★ Supplementary Counsel :-

$\\\;\sf{\leadsto\;\;For\;imaginary\;roots\;:\;b^{2}\;-\;4ac\;<\;0}$

$\\\;\sf{\leadsto\;\;For\;distinct\;roots\;:\;b^{2}\;-\;4ac\;>\;0}$

Answer 2

Answer:

\begin{gathered}\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}\end{gathered}

UnderstandingtheQuestion:−

Here the Concept of Quadratic Equations has been used. Here we see that we are given a quadratic equation. We know that a Quadratic Equation has 2 roots either can be equal, imaginary or distinct but here we have to go with equal case since, we need to find the value of a such that there are only one roots that is only one value of x.

Let's do it !!

_______________________________________________

★ Formula Used :-

\begin{gathered}\\\;\boxed{\sf{For\;equal\;roots\;:\;b^{2}\;-\;4ac\;=\;\bf{0}}}\end{gathered}

Forequalroots:b

2

−4ac=0

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★ Solution :-

Given,

✒ ax² - 5x + 2 = 0

Here, a = a, b = -5 and c = 2.

We know that,

\begin{gathered}\\\;\;\sf{:\mapsto\;\;For\;equal\;roots\;:\;b^{2}\;-\;4ac\;=\;\bf{0}}\end{gathered}

:↦Forequalroots:b

2

−4ac=0

\begin{gathered}\\\;\;\sf{:\mapsto\;\;b^{2}\;-\;4ac\;=\;\bf{0}}\end{gathered}

:↦b

2

−4ac=0

By applying values, we get,

\begin{gathered}\\\;\;\sf{:\mapsto\;\;(-\:5)^{2}\;-\;4\;\times\;a\;\times\;2\;=\;\bf{0}}\end{gathered}

:↦(−5)

2

−4×a×2=0

\begin{gathered}\\\;\;\sf{:\mapsto\;\;25\;-\;8\;\times\;a\;=\;\bf{0}}\end{gathered}

:↦25−8×a=0

\begin{gathered}\\\;\;\sf{:\mapsto\;\;\;-\;8\;\times\;a\;=\;\bf{-\:25}}\end{gathered}

:↦−8×a=−25

\begin{gathered}\\\;\;\sf{:\mapsto\;\;\;8\;\times\;a\;=\;\bf{25}}\end{gathered}

:↦8×a=25

\begin{gathered}\\\;\;\underline{\underline{\bf{:\mapsto\;\;\;a\;=\;\bf{\dfrac{25}{8}}}}}\end{gathered}

:↦a=

8

25

Also, when a = 0 , we get,

✒ ax² - 5x + 2 = 0

✒ 0x² - 5x + 2 = 0

✒ -5x = -2

✒ 5x = 2

✒ x = 2/5

This also, gives single root of equation since it forms linear equation. Thus,

a = 0

is also correct answer.

\begin{gathered}\\\;\underline{\boxed{\tt{Hence\;\;for\;\;single\;\;roots,\;a\;=\;\bf{\dfrac{25}{8}\;\;\tt{or}\;\;\bf{0}}}}}\end{gathered}

Henceforsingleroots,a=

8

25

or0

_______________________________________________

★ More to know :-

• Verification ::

We need to verify, if the equation has single roots or not for the value of a we got. Then,

✒ ax² - 5x + 2 = 0

✒ (25/8)x² - 5x + 2 = 0

✒ 25x² - 40x + 8 = 0

Now using the Quadratic Formula, we get,

\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:b\;\pm\;\sqrt{b^{2}\;-\;4ac}}{2a}}\end{gathered}

⇒x=

2a

−b±

b

2

−4ac

\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:40\;\pm\;\sqrt{(-\:40)^{2}\;-\;4\;\times\;25\;\times\;8}}{2\;\times\;25}}\end{gathered}

⇒x=

2×25

−40±

(−40)

2

−4×25×8

\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:(-\:40)\;\pm\;\sqrt{1600\;-\;1600}}{50}}\end{gathered}

⇒x=

50

−(−40)±

1600−1600

\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{40\;\pm\;\sqrt{0}}{50}}\end{gathered}

⇒x=

50

40±

0

\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{40}{50}\;=\;\dfrac{4}{5}}\end{gathered}

⇒x=

50

40

=

5

4

Clearly,

\begin{gathered}\\\;\bf{\Rightarrow\;\;x\;=\;\dfrac{4}{5}\;or\;\dfrac{4}{5}}\end{gathered}

⇒x=

5

4

or

5

4

This is because, this is Quadratic equation so it must have two equal roots.

This gives that this equation has only single root that is 4/5 when a = 25/8

So our answer is correct.

Hence, Verified.

_______________________________________________

★ Supplementary Counsel :-

\begin{gathered}\\\;\sf{\leadsto\;\;For\;imaginary\;roots\;:\;b^{2}\;-\;4ac\; < \;0}\end{gathered}

⇝Forimaginaryroots:b

2

−4ac<0

\begin{gathered}\\\;\sf{\leadsto\;\;For\;distinct\;roots\;:\;b^{2}\;-\;4ac\; > \;0}\end{gathered}

⇝Fordistinctroots:b

2

−4ac>0